已知|x|≤1,|y|≤1,证明|x+y|≤|1+xy|

证明：可以利用作差比较法∵ |1+xy|²-|x+y|²= (1+xy)²-(x+y)²=1+2xy+x²y²-(x²+2xy+y²)=1+x²y²-x²-y²=1-x²+y²(x²-1)=(1-x²)(1-y&#...不好意思，要用分析法证明一样啊，倒过来即可要证 |x+y|≤|1+xy|只需证|1+xy|²≥|x+y|²只需证 1+2xy+x²y²≥x²+2xy+y²只需证1+x²y²≥x²+y²只需证1+x²y²-x²-y²≥0只需证 (1-x²)(1-y²)≥0∵|x|≤1,|y|≤1∴1-x²≥0,1-y²≥0∴(1-x²)(1-y²)≥0成立∴ |x+y|≤|1+xy|